AQA AS Level Physics Paper 2, June 2025: Question 29

1 mark · Medium difficulty · Multiple Choice

Determine which cylindrical rod with diameter 2d has the same length as a rod R of density ρ, mass m, and diameter d.

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Question

Question 29: A cylindrical rod R is made of a material with density rho, mass m, and diameter d. Four other cylindrical rods A, B, C, and D each have diameter 2d. A table shows their properties: Rod A has density 2 rho and mass 2m; Rod B has density 2 rho and mass 4m; Rod C has density 2 rho and mass 8m; Rod D has density 4 rho and mass 4m. Students must select which rod has the same length as R.

Mark scheme

Show the mark scheme Mark scheme for question 29 shows the correct answer is C, with density 2 rho and mass 8m, categorized under assessment objective AO1.

How to answer it

Comparing Cylindrical Rod Dimensions Using Density

📌 What this question tests

This multiple-choice question assesses your ability to manipulate the relationship between density, mass, and volume ( ρ = m / V ) alongside geometric scaling for a 3D cylinder. Specifically, it tests your understanding of how changes in diameter scale cross-sectional area ( A ∝ d² ) and how to equate expressions for length across different materials and dimensions.

Question 29 • Multiple Choice [1 Mark]

Determining Which Rod Has the Same Length as Rod R

AQA AS Physics • Matter and Radiation • AO1

📐 Step-by-Step Derivation

  1. Formula for the base rod R:
    Volume of a cylinder is V = A × L , where A = π(d/2)² = (πd²)/4 .
    Density is defined as:
    ρ = m / V = m / (A × L)
    Rearranging for length of rod R:
    L = m / (ρ × A)
  2. Determine the new cross-sectional area (A'):
    Each new rod has diameter 2d .
    A' = π(2d / 2)² = 4 × [π(d/2)²] = 4A
    Doubling the diameter quadruples the cross-sectional area ( 2² = 4 ).
  3. Set up the condition for equal length ( L' = L ):
    For any new rod with density ρ' and mass m' :
    L' = m' / (ρ' × A') = m' / (ρ' × 4A)
    For L' = L :
    m' / (4 × ρ' × A) = m / (ρ × A)
    Cancel out common factor A :
    m' = 4 × (ρ' / ρ) × m
  4. Test the given options:
    • For A: ρ' = 2ρ ⇒ requires m' = 4 × 2 × m = 8m . Given mass is 2m (incorrect; L' = 0.25 L ).
    • For B: ρ' = 2ρ ⇒ requires m' = 8m . Given mass is 4m (incorrect; L' = 0.5 L ).
    • For C: ρ' = 2ρ and mass is 8m ⇒ L' = 8m / (2ρ × 4A) = 8m / 8ρA = m / (ρA) = L (Correct!)
    • For D: ρ' = 4ρ ⇒ requires m' = 4 × 4 × m = 16m . Given mass is 4m (incorrect; L' = 0.25 L ).

✅ Correct Answer

Option C (2ρ, 8m)

Mark Scheme: Award 1 mark for identifying option C [AO1].

💡 Key Knowledge

  • Density: ρ = m / V (SI unit: kg m⁻³ ).
  • Cylinder Volume: V = πr²L = (πd²L)/4 .
  • Scaling Factor Rule: Scaling linear dimension d by factor k multiplies area by k² and volume by k³ (if scaled proportionally in all directions).

🧠 Exam Technique

  • Proportional reasoning: Write a proportionality relation before looking at numbers: L ∝ m / (ρ × d²) .
  • To keep L constant, the numerator and denominator must change by the exact same scale factor:
    scale factor of m = scale factor of (ρ × d²) .
  • Here, d² increases by 2² = 4 . For density factor 2 , total denominator factor is 2 × 4 = 8 . Therefore, mass must increase by 8 !

❌ Common Errors

  • Forgetting to square the diameter: Assuming doubling diameter doubles the area ( 2× instead of 4× ). This leads to wrongly selecting B ( 2 × 2 = 4 ).
  • Inverting the ratio: Dividing mass by volume upside down or incorrectly rearranging V = m / ρ .
  • Ignoring the question's premise: Missing the detail in the question stem that all rods A–D have diameter 2d .

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.